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Find A and B so that F(x) is a Differentiable Function

  1. Mar 1, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the values of a and b that make f a differentiable function.

    Note: F(x) is a piecewise function

    f(x):
    Ax^2 - Bx, X ≤ 1
    Alnx + B, X > 1
    2. Relevant equations


    3. The attempt at a solution
    Made the two equations equal each other.
    Ax^2 - Bx = Alnx + B
    Inserting x=1 gives,
    A - B = B, which is also A - 2B = 0, which also means A = 2B

    Deriving the equation,
    2Ax - B = A/X
    Inserting x=1 here gives,
    2A - B = A, which is also A - B = 0, which also means A = B

    By then I'm stumped here.
    I try to eliminate either A and B with,
    A - B = 0
    A - 2B = 0
    In the end, both A and B would have to equal zero, both of which doesn't work.

    If I have A = 2B then,
    F'(x): A - B = 0 ⇒ 2B - B = 0 ⇒ B = 0
    As said before, B = 0 would not be the right answer, as far as I know at least.

    If A = B, then
    F(x): A - 2B = 0 ⇒ B - 2B = 0 ⇒ B = 0
    Once Again, B equaling zero would not work.

    Right now, i'm convinced this problem is virtually impossible.
     
  2. jcsd
  3. Mar 1, 2017 #2
    Are you sure you copied the problem down correctly?
     
  4. Mar 1, 2017 #3
    Yep, that's exactly what it says.
     
  5. Mar 1, 2017 #4
    Then I'm stumped too. Perhaps someone else will have a solution.
     
  6. Mar 1, 2017 #5
  7. Mar 2, 2017 #6

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    So what is wrong with ##f(x) \equiv 0##? That's a perfectly good differentiable function.
     
  8. Mar 2, 2017 #7
    UPDATE:

    My Math Teacher just said it was a typo, where he forgot to add a negative two, so that the piecewise function would be:

    f(x):
    Ax^2 - Bx - 2, X ≤ 1
    Alnx + B, X > 1

    Now solving it, I got A = -2 and B = -2.
     
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